Question: Simplify; express your answer in exponential form. Assume $y\neq 0, r\neq 0$. $\dfrac{{(y^{-4})^{-3}}}{{(y^{5}r^{-4})^{-2}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${y^{-4}}$ to the exponent ${-3}$ . Now ${-4 \times -3 = 12}$ , so ${(y^{-4})^{-3} = y^{12}}$ In the denominator, we can use the distributive property of exponents. ${(y^{5}r^{-4})^{-2} = (y^{5})^{-2}(r^{-4})^{-2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(y^{-4})^{-3}}}{{(y^{5}r^{-4})^{-2}}} = \dfrac{{y^{12}}}{{y^{-10}r^{8}}}$ Break up the equation by variable and simplify. $\dfrac{{y^{12}}}{{y^{-10}r^{8}}} = \dfrac{{y^{12}}}{{y^{-10}}} \cdot \dfrac{{1}}{{r^{8}}} = y^{{12} - {(-10)}} \cdot r^{- {8}} = y^{22}r^{-8}$.